1 Answer
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Re 2, for $n=1$, $P_{ij}=f^1_{ij}$ for every $i\ne j$ and $1-P_{ii}$ is the sum of $f^1_{ij}$ over $j\ne i$, hence one recovers trivially $P$ from $f^1=(f^1_{ij})_{ij}$.

Re 2 again, on the contrary, there is no hope to recover $P$ from $f^n=(f^n_{ij})_{ij}$ in general, for any given $n\ge2$. For example, the two (deterministic) one-step rotations on the discrete circle with $2n$ vertices, clockwise and counterclockwide, yield the same matrix $f^n$ although their transition matrices $P$ are different.